
Result when two soluble compounds mix and
an insoluble (solid) compound is produced.
AgNO3 (aq) + NaCl (aq) 
AgCl (s) + NaNO3 (aq)
This is the insoluble compound
(precipitate). Precipitation is
established by looking at the
solubility table. Insoluble
compounds are designated (s) for
solid.
This compound
is soluble. The
(aq) designates
that it dissolves
in water.

We predict the products by matching the
cation of one reactant to the anion in the
other reactant, and vice versa.
silver cation + chloride anion
AgNO3 (aq) + NaCl (aq) 
nitrate anion + sodium cation
AgCl (s) + NaNO3 (aq)

Step 1. Check to see if a precipitate forms
using the solubility table.
this is insoluble
Ca(NO3)2 (aq) + NaF (aq) 
this is soluble

Step 2. Write the products with cations and
anions in a 1:1 ratio.
this is insoluble
Ca(NO3)2 (aq) + NaF (aq)  CaF (s) + NaNO3 (aq)
this is soluble
One calcium ion to one
fluoride ion. The (s) is added
because this is the insoluble
compound.
One sodium to one
nitrate ion. Nitrate is a
polyatomic ion; we do
not alter it’s formula.
The (aq) is used for the
soluble compounds.

Step 3. Determine the charges of each ion on
the product side, and balance the products
using subscripts.
+2
-1
2
+1
-1
Ca(NO3)2 (aq) + NaF (aq)  CaF2 (s) + NaNO3 (aq)
Since these charges did not
balance out, a subscript of 2
was added to fluoride so the
total charges would add to zero.
Since these
charges do
balance out,
no subscript
is needed.

Step 4. Balance the equation using
coefficients.
+2
-1
2
+1
-1
Ca(NO3)2 (aq) + 2 NaF (aq)  CaF2 (s) + 2 NaNO3 (aq)
These are
added to
balance the
equation.


The total ionic equations represents what
each compound looks like when dissolved in
water.
Strong electrolytes break apart 100% in water to form ions, so
these are shown as ions in the equation.
◦ NaCl (aq)  Na+ (aq) + Cl- (aq)

Weak electrolytes break apart ~5% in water, so the
predominant form of these species is the neutral molecule.
◦ HF (aq) is shown as HF (aq)

Non-electrolytes do not break apart at all. Like weak
electrolytes, they are shown as is.
H2O (l) is shown as H2O (l)

The bottom line: Separate into ions everything
with an (aq) except weak acids.
This is a solid; it will not break apart.
Ca(NO3)2 (aq) + 2 NaF (aq)  CaF2 (s) + 2 NaNO3 (aq)
These are soluble and will split into ions.

The bottom line: Separate into ions everything
with an (aq) except weak acids.
This subscripts will become a coefficient. It is not
part of the polyatomic ion’s formula.
Ca(NO3)2 (aq) + 2 NaF (aq)  CaF2 (s) + 2 NaNO3 (aq)
These subscripts will not be
moved because they are part
of the polyatomic ion.

The bottom line: Separate into ions everything
with an (aq) except weak acids.
Ca(NO3)2 (aq) + 2 NaF (aq)  CaF2 (s) + 2 NaNO3 (aq)
The total ionic equation is:
Ca2+(aq) + 2 NO3-(aq) + 2 Na+(aq) + 2 F-(aq)  CaF2(s) +
2 Na+(aq) + 2 NO3-(aq)

Eliminate the spectator ions from the total
ionic equation. The spectator ions are the
species that look identical in every respect on
both sides of the equation.
Ca2+(aq) + 2 NO3-(aq) + 2 Na+(aq) + 2 F-(aq)  CaF2(s) +
2 Na+(aq) + 2 NO3-(aq)
The net ionic equation is:
Ca2+(aq)+ 2 F-(aq)  CaF2(s)

The formula unit equation
Ca(NO3)2 (aq) + 2 NaF (aq)  CaF2 (s) + 2 NaNO3 (aq)

The total ionic equation
Ca2+(aq) + 2 NO3-(aq) + 2 Na+(aq) + 2 F-(aq)  CaF2(s) +
2 Na+(aq) + 2 NO3-(aq)

The net ionic equation
Ca2+(aq)+ 2 F-(aq)  CaF2(s)